November 13th - pg.338

Page 338.

7.

NH3(g) + HNO3(l) à NH4NO3(s)

a) Reactants

H°f(NH3) = -45.9 kJ
H°f(HNO3) = -174.1 kJ

Products

H°f(NH4NO3) = -365.6 kJ

H = H°f (products) - H°f (reactants)
= -365.6 – [-45.9 + (-174.1)]
= -365.6 – (-220)
= -365.6 + 220
= -145.6 kJ

b)



c)
50T = 50000kg = 50000000g of NH4NO3

1 mol of NH4NO3 = 80.06g
X mols = 50000000g
= 624531.6 mols

-145.6 kJ = x kJ
Mol 624531.6 mol
= - 9.1 x 107


8.
Enthalpy for the formation of anthracite (C52H16O) = -396.4 kJ

Anthracite Coal à 100kg = 100000g

2 C52H16O(s) + 111O2(g) à 104CO2(g) + 16H2O

Change in H = Sum of Hºf(products) – Sum of Hºf (reactants)
= [(104)(-393.5) + (16)(-241.8)] – [(2)(_396.4) + (111)(0)]
= (-40924 – 3868.8) – (-792.8 + 0)
= (-44792.8 + 792.8)
= -44, 000 kJ

1 mol anthracite = 656.68g
x mols = 100000g
= 152.28mol

1 mol = -22000kJ
152.28 = x kJ
x = -3350160kJ
= -3.4 x 106kJ
= -3.34 x 103MJ